Two slits are kept 1 mm apart and the screen is placed 1.0 m away. What is the fringe separation when blue-green light of wavelength 500 nm is used?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.5 mm
In Young's double-slit experiment the fringe separation (fringe width) is \( \beta = \lambda D/d \), where \( D \) is the slit-screen distance and \( d \) the slit separation.
\( \lambda = 500\,\mathrm{nm} = 5\times10^{-7}\,\mathrm{m} \)
\( d = 1\,\mathrm{mm} = 1\times10^{-3}\,\mathrm{m} \)
\( D = 1.0\,\mathrm{m} \)
\( \displaystyle \beta = \frac{\lambda D}{d} = \frac{5\times10^{-7} \times 1.0}{1\times10^{-3}} = 5\times10^{-4}\,\mathrm{m} = 0.5\,\mathrm{mm} \)
Note: A similar question was also asked in CUET (UG) 2025 on 29-05-2025 (Slot 2), and in CUET (UG) 2023 on 13-06-2023 (Slot 3).
Note: This question is from the NCERT Exemplar. NTA sets numerical questions from the Exemplar too, so do not ignore it while practising numericals.
NCERT Exemplar:
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